1974 AMC 12 Problem 18

Attempt Problem 18 of the 1974 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1974 AMC 12 solutions, or check the answer key.

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18.

\log_8 3=p

and

\log_3 5=q,

then, in terms of

p

and

q,

\log_{10}5

equals

pq

\frac{3p+q}{5}

\frac{1+3pq}{p+q}

\frac{3pq}{1+3pq}

p^2+q^2

Convert every logarithm to base

2

The givens imply

\log_2 3=3p

and

\log_2 5=3pq

From

p=\log_8 3,

\log_2 3=3p.

Therefore

\log_2 5=(\log_2 3)(\log_3 5)=3pq.

Since

\log_2 10=1+\log_2 5=1+3pq,

\log_{10}5=\frac{\log_2 5}{\log_2 10} =\frac{3pq}{1+3pq}.

Therefore, the correct answer is D.

Problem 18 in other years